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    Let fn(x)=(2 + (−2)n ) x2 + (n+3)x + n2 where n is a positive integer and x is any real number.
    Find an expression, simplified as much as possible, for f1 (f1 (f1 (··· f1 (x)))) where f1 is applied k times.
    [Here k is a positive integer.]

    Answer is...

    More generally f(1 k times)(x) = 4k x + 1 + 4 + · · · + 4k−1= 4k x +(4^k − 1)/3

    How do you work this answer out (The '(4^k -1)/3' part of the answer).

    Basically, how do you get from (4k-1 + 4k-2 + ... + 4k-k)

    to (4^k -1)/3
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    (Original post by KloppOClock)
    Let fn(x)=(2 + (−2)n ) x2 + (n+3)x + n2 where n is a positive integer and x is any real number.
    Find an expression, simplified as much as possible, for f1 (f1 (f1 (··· f1 (x)))) where f1 is applied k times.
    [Here k is a positive integer.]

    Answer is...

    More generally f(1 k times)(x) = 4k x + 1 + 4 + · · · + 4k−1= 4k x +(4^k − 1)/3

    How do you work this answer out (The '(4^k -1)/3' part of the answer).

    Basically, how do you get from (4k-1 + 4k-2 + ... + 4k-k)

    to (4^k -1)/3
    Are you sure it's not \frac{4}{3}(4^k-1) instead??
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    (Original post by RDKGames)
    Are you sure it's not \frac{4}{3}(4^k-1) instead??
    https://www.maths.ox.ac.uk/system/fi...olutions07.pdf

    question 2 part ii
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    Ah I see, never mind, was thinking slightly differently.

    Okay you have:

    4^0+4^1+4^2+...+4^k

    This is a geometric sequence and you need to use a formula to sum it up where S_n=\frac{a(1-r^n)}{1-r} where a is the first term and r is the common ratio.
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    (Original post by RDKGames)
    Ah I see, never mind, was thinking slightly differently.

    Okay you have:

    4^0+4^1+4^2+...+4^k

    This is a geometric sequence and you need to use a formula to sum it up where S_n=\frac{a(1-r^n)}{1-r} where a is the first term and r is the common ratio.
    ahh right, i forgot all about sequences and series. thank you.
 
 
 
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